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Disintegration of a particle

  1. Nov 3, 2007 #1
    First, sorry for my English. I'm not very well in it... Please, try to understand :)
    The problem.
    We have reaction: [tex] \mu \rightarrow e + \nu + \tilde{\nu}[/tex]
    We know energy of myuon - E.
    Question: Find the maximum and the minimum energy of electron.

    My attept:

    Conservation of energy: [tex] E = E_e+E_{\nu}+E_{\tilde{\nu}}[/tex] (1)
    Conservation of impulse: [tex] \vec{p} = \vec{p_e}+\vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}}[/tex] (2)

    The mass of the rest of neutrino and antineutrino is 0. So, [tex]E_{\nu}=p_{\nu}c[/tex]
    [tex]E_{\tilde{\nu}}=p_{\tilde{\nu}}c[/tex] and from the first equation:

    [tex] E_e = E-c(p_{\nu}+p_{\tilde{\nu}})[/tex]

    Therefore, we must find the minumum and the maximum value of [tex](p_{\nu}+p_{\tilde{\nu}})[/tex] Then, the minmum value of this expression gives us the maximum value of [tex]E_e[/tex] and the maximum value gives the minimum of energy. Am I right in this statement?

    From (2):
    [tex] \vec{p_{\nu}}+\vec{p_{\tilde{\nu}}}} = \vec{p} - \vec{p_e} [/tex]
    Also we know that: [tex]p= \sqrt{\frac{E^2}{c^2}-m^2c^2}[/tex]
    and: [tex]p_e= \sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}[/tex]

    How can I find the minumum and the maximum value of [tex](p_{\nu}+p_{\tilde{\nu}})[/tex] with the help of all I wrote here? :)

    I also have an assumption that we can find half of the answer simply in the following way:
    [tex]E_e[/tex] reaches its maximum when impulses of neutrino and antineutrino have opposite directions and equals in absolute. (It's easy to understand this fact because in this case impulses of neutrino and antineutrino compensate each other and the value [tex]p_e[/tex] reaches its maximum, so does [tex]E_e[/tex]). Then, almost easy:
    [tex]p=\sqrt{\frac{E^2}{c^2}-m^2c^2}=p_e=\sqrt{\frac{{E_e}^2}{c^2}-{m_e}^2c^2}[/tex]
    And we have: [tex]E_e^{max}=\sqrt{E^2+c^4({m_e}^2-m^2)}[/tex] Am I right? How can I find the minimum value of [tex]E_e[/tex]?

    Please, help.
     
    Last edited: Nov 3, 2007
  2. jcsd
  3. Nov 3, 2007 #2

    arivero

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    Gold Member

    No issue with the language, except that I am curious about what country considers beta or weak decay to be "Introductory Physics"
     
    Last edited: Nov 3, 2007
  4. Nov 3, 2007 #3

    malawi_glenn

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    Science Advisor
    Homework Helper

    sweden do =)

    And also, there are guys posting mass falling to the earth without corilos and centrifugal terms in the advanced forum. Very hard to get everything right here..
     
  5. Nov 3, 2007 #4
    I'm very sorry, what forum corresponds to this theme?
     
  6. Nov 3, 2007 #5

    Astronuc

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    Staff: Mentor

    Well, it's not a problem really, but this topic is appropriate for the Advanced Physics HW forum. This is upper level material as opposed to introductory physiscs.
     
  7. Nov 3, 2007 #6
    Please, delete this theme. I posted this problem in "Advenced Physics".
     
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